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Question Number 197766 by universe last updated on 28/Sep/23

∫_0 ^(π/2) (lim_(n→∞) nsin^(2n+1) x cos x)dx = ?

$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\mathrm{sin}^{\mathrm{2}{n}+\mathrm{1}} {x}\:\mathrm{cos}\:{x}\right){dx}\:\:=\:? \\ $$

Commented by witcher3 last updated on 28/Sep/23

dominate cv Theorem

$$\mathrm{dominate}\:\mathrm{cv}\:\mathrm{Theorem} \\ $$

Answered by witcher3 last updated on 28/Sep/23

=∫_0 ^1 lim_(n→∞) x^(2n+1) dx,sin(x)→x lim_(n→∞) ∫_0 ^1 nx^(2n+1) dx=lim_(n→∞) (n/(2n+2))=(1/2) x^n =u⇒dx=nx^(n−1) dx=du x=u^(1/n) =lim_(n→∞) ∫_0 ^1 u.u^(2/n) du=lim_(n→∞) ∫_0 ^1 u^(1+(2/n)) ≤lim_(n→∞) ∫_0 ^1 du=1 lim_(n→∞) ∫_0 ^1 nsin^(2n+1) (x)cos(x)dx=(1/2)=∫_0 ^1 lim_(n→∞) nsin^(2n+1) (x)cos(x)dx

$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}x}^{\mathrm{2n}+\mathrm{1}} \mathrm{dx},\mathrm{sin}\left(\mathrm{x}\right)\rightarrow\mathrm{x} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{nx}^{\mathrm{2n}+\mathrm{1}} \mathrm{dx}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{n}}{\mathrm{2n}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{n}} =\mathrm{u}\Rightarrow\mathrm{dx}=\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \mathrm{dx}=\mathrm{du} \\ $$$$\mathrm{x}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}.\mathrm{u}^{\frac{\mathrm{2}}{\mathrm{n}}} \mathrm{du}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}^{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{n}}} \leqslant\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{du}=\mathrm{1} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{nsin}^{\mathrm{2n}+\mathrm{1}} \left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}nsin}^{\mathrm{2n}+\mathrm{1}} \left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$ \\ $$

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